
This post is about a basic fact which is not very well known. Let (μi)i∈I be a family of probability measures on a topological space E equipped with its Borel sigma-field. The following two properties are equivalent, and when they hold we say that (μi)i∈I is tight.
- for any ε>0 there exists a compact set Kε⊂E such that
sup
- there exists a measurable {f:E\rightarrow[0,\infty]} with compact sub-level-sets such that
\sup_{i\in I}\int\!f\,d\mu_i<\infty.
Recall that the sub-level-sets of {f} are the sets {\{x\in E:f(x)\leq r\}} , {r\geq0} . If {E} is not compact then it cannot be a sub-level-set of {f} and thus {f} cannot be bounded, while if {E} is compact then the property holds trivially with {f} constant.
To deduce 1. from 2. we write, using the Markov inequality, for any {r>0} ,
\sup_{i\in I}\mu_i(\{x\in E:f(x)\leq r\}^c) \leq \frac{1}{r}\sup_{i\in I}\int\!f\,d\mu_i=\frac{C}{r},
which leads to take {r=r_\varepsilon=C/\varepsilon} and {K_\varepsilon=\{x\in E:f(x)\leq r_\varepsilon\}} , for any {\varepsilon>0} .
To deduce 2. from 1. we first extract from 1. a sequence of compact subsets {{(K_{1/n^2})}_{n\geq1}} . We can assume without loss of generality that it grows: {K_{1/n^2}\subset K_{1/m^2}} if {n\leq m} . Now, for any {x\in F=\cup_{n}K_n} , there exists {n_x} such that {x\in K_{1/m^2}} for any {m\geq n_x} , and thus {\sum_n \mathbf{1}_{K_{1/n^2}^c}(x)=n_x-1<\infty} . As a consequence, if one defines
f:=\sum_n\mathbf{1}_{K_{1/n^2}^c}
then {f<\infty} on {F} while {f=\infty} on {F^c} , and {f} has compact sub-level-sets since {\{x\in E:f(x)\leq n-1\}=K_{1/n^2}} for any {n\geq1} . On the other hand, by definition of {K_\varepsilon} ,
\sup_{i\in I}\int\!f\,d\mu_i \leq\sum_n\frac{1}{n^2}<\infty.
Tightness is an important concept of probability theory. A famous theorem of Prokhorov states that a family of probability measures is tight if and only if it is relatively compact for the topology of narrow convergence.
Taking {X_i\sim\mu_i} for every {i\in I} , the second property reads
\sup_{i\in I}\mathbb{E}(f(X_i))<\infty.
It plays for tightness the role played by the famous de la Vallée Poussin criterion for uniform integrability.
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